Electric Charge
Properties of Electric Charge
• There are two kinds of electric charge.
– like charges repel
– unlike charges attract
• Electric charge is conserved.
– Positively charged particles are called protons.
– Uncharged particles are called neutrons.
– Negatively charged particles are called electrons.
Electric Charge
Properties of Electric Charge, continued
• Electric charge is quantized. That is, when an object
is charged, its charge is always a multiple of a
fundamental unit of charge.
• Charge is measured in coulombs (C).
• The fundamental unit of charge, e, is the magnitude
of the charge of a single electron or proton.
e = 1.602 176 x 10–19 C
Electric Charge
Transfer of Electric Charge
• An electrical conductor is a material in which
charges can move freely.
• An electrical insulator is a material in which charges
cannot move freely.
Electric Charge
Transfer of Electric Charge, continued
• Insulators and conductors can be charged by contact.
• Conductors can be charged by induction.
• Induction is a process of charging a conductor by
bringing it near another charged object and
grounding the conductor.
Electric Charge
Transfer of Electric Charge, continued
• A surface charge can be
induced on insulators by
polarization.
• With polarization, the
charges within individual
molecules are realigned
such that the molecule
has a slight charge
separation.
Electric Force
Coulomb’s Law
• Two charges near one another exert a force on one
another called the electric force.
• Coulomb’s law states that the electric force is proportional to the magnitude of each charge and inversely
proportional to the square of the distance between
them.
qq 
Felectric  kC  1 2 2 
 r 
electric force = Coulomb constant 
 charge 1 charge 2 
2
distance


Electric Force
Coulomb’s Law, continued
• The resultant force on a charge is the vector sum of
the individual forces on that charge.
• Adding forces this way is an example of the principle
of superposition.
• When a body is in equilibrium, the net external force
acting on that body is zero.
Electric Force
Sample Problem
The Superposition Principle
Consider three point
charges at the corners of a
triangle, as shown at right,
where q1 = 6.00  10–9 C,
q2 = –2.00  10–9 C, and
q3 = 5.00  10–9 C. Find
the magnitude and
direction of the resultant
force on q3.
Electric Force
Sample Problem, continued
The Superposition Principle
1. Define the problem, and identify the known
variables.
Given:
q1 = +6.00  10–9 C
r2,1 = 3.00 m
q2 = –2.00  10–9 C
r3,2 = 4.00 m
q3 = +5.00  10–9 C
r3,1 = 5.00 m
q = 37.0º
Unknown: F3,tot = ? Diagram:
Electric Force
Sample Problem, continued
The Superposition Principle
Tip: According to the superposition principle, the resultant
force on the charge q3 is the vector sum of the forces
exerted by q1 and q2 on q3. First, find the force exerted on
q3 by each, and then add these two forces together
vectorially to get the resultant force on q3.
2. Determine the direction of the forces by analyzing
the charges.
The force F3,1 is repulsive because q1 and q3 have
the same sign.
The force F3,2 is attractive because q2 and q3 have
opposite signs.
Electric Force
Sample Problem, continued
The Superposition Principle
3. Calculate the magnitudes of the forces with
Coulomb’s law.


2  5.00  10 –9 C 6.00  10 –9 C


q3q1
N

m
9
F3,1  kC
  8.99  10
 
2
2
2
(r 3,1)
C
 5.00 m 


 


F3,1  1.08  10 –8 N
F3,2


2  5.00  10 –9 C 2.00  10 –9 C

q3q2
9 Nm 
 kC
  8.99  10
 
2
2
2
(r 3,2)
C 
4.00m




F3,1  5.62  10 –9 N
 


Electric Force
Sample Problem, continued
The Superposition Principle
4. Find the x and y components of each force.
At this point, the direction each component must be
taken into account.
F3,1: Fx = (F3,1)(cos 37.0º) = (1.08  10–8 N)(cos 37.0º)
Fx = 8.63  10–9 N
Fy = (F3,1)(sin 37.0º) = (1.08  10–8 N)(sin 37.0º)
Fy = 6.50  10–9 N
F3,2: Fx = –F3,2 = –5.62  10–9 N
Fy = 0 N
Electric Force
Sample Problem, continued
The Superposition Principle
5. Calculate the magnitude of the total force acting
in both directions.
Fx,tot = 8.63  10–9 N – 5.62  10–9 N = 3.01  10–9 N
Fy,tot = 6.50  10–9 N + 0 N = 6.50  10–9 N
Electric Force
Sample Problem, continued
The Superposition Principle
6. Use the Pythagorean theorem to find the magnitude of the resultant force.
F3,tot  (Fx ,tot )2  (Fy ,tot )2  (3.01 109 N)2  (6.50  109 N)2
F3,tot  7.16  10 –9 N
Electric Force
Sample Problem, continued
The Superposition Principle
7. Use a suitable trigonometric function to find the
direction of the resultant force.
In this case, you can use the inverse tangent function:
tan  
Fy ,tot
Fx ,tot
  65.2º
6.50  10 –9 N

3.01 10 –9 N
Electric Force
Coulomb’s Law, continued
• The Coulomb force is a field force.
• A field force is a force that is exerted by one object on
another even though there is no physical contact
between the two objects.
The Electric Field
Electric Field Strength
• An electric field is a region where an electric force
on a test charge can be detected.
• The SI units of the electric field, E, are newtons per
coulomb (N/C).
• The direction of the electric field vector, E, is in the
direction of the electric force that would be exerted on
a small positive test charge.
The Electric Field
Electric Field Strength, continued
• Electric field strength depends on charge and
distance. An electric field exists in the region around
a charged object.
• Electric Field Strength Due to a Point Charge
E  kC
q
r2
electric field strength = Coulomb constant 
charge producing the field
 distance 
2
The Electric Field
Sample Problem
Electric Field Strength
A charge q1 = +7.00 µC is
at the origin, and a charge
q2 = –5.00 µC is on the xaxis 0.300 m from the
origin, as shown at right.
Find the electric field
strength at point P,which is
on the y-axis 0.400 m from
the origin.
The Electric Field
Sample Problem, continued
Electric Field Strength
1. Define the problem, and identify the known
variables.
Given:
q1 = +7.00 µC = 7.00  10–6 C
r1 = 0.400 m
q2 = –5.00 µC = –5.00  10–6 C
r2 = 0.500 m
q = 53.1º
Unknown:
E at P (y = 0.400 m)
Tip: Apply the principle of
superposition. You must first
calculate the electric field produced
by each charge individually at point
P and then add these fields
together as vectors.
The Electric Field
Sample Problem, continued
Electric Field Strength
2. Calculate the electric field strength produced by
each charge. Because we are finding the magnitude
of the electric field, we can neglect the sign of each
charge.
–6
q1
9
2
2  7.00  10 C 
5
E1  kC 2  8.99  10 N  m /C 

3.93

10
N/C
2 
r1
 (0.400 m) 


–6
q2
9
2
2  5.00  10 C 
5
E2  kC 2  8.99  10 N  m /C 

1.80

10
N/C
2 
r2
 (0.500 m) 


The Electric Field
Sample Problem, continued
Electric Field Strength
3. Analyze the signs of the
charges.
The field vector E1 at P due
to q1 is directed vertically
upward, as shown in the
figure, because q1 is
positive. Likewise, the field
vector E2 at P due to q2 is
directed toward q2 because
q2 is negative.
The Electric Field
Sample Problem, continued
Electric Field Strength
4. Find the x and y components of each electric field
vector.
For E1: Ex,1 = 0 N/C
Ey,1 = 3.93  105 N/C
For E2: Ex,2= (1.80  105 N/C)(cos 53.1º) = 1.08  105 N/C
Ey,1= (1.80  105 N/C)(sin 53.1º)= –1.44  105 N/C
The Electric Field
Sample Problem, continued
Electric Field Strength
5. Calculate the total electric field strength in both
directions.
Ex,tot = Ex,1 + Ex,2 = 0 N/C + 1.08  105 N/C
= 1.08  105 N/C
Ey,tot = Ey,1 + Ey,2 = 3.93  105 N/C – 1.44  105 N/C
= 2.49  105 N/C
The Electric Field
Sample Problem, continued
Electric Field Strength
6. Use the Pythagorean theorem to find the
magnitude of the resultant electric field strength
vector.
Etot 
E   E 
Etot 
1.08  10 N/C    2.49  10 N/C 
2
x ,tot
2
y ,tot
5
Etot  2.71 105 N/C
2
5
2
The Electric Field
Sample Problem, continued
Electric Field Strength
7. Use a suitable trigonometric function to find the
direction of the resultant electric field strength
vector.
In this case, you can use the inverse tangent
function:
tan  
E y ,tot
E x ,tot
  66.0
2.49  105 N/C

1.08  105 N/C
The Electric Field
Sample Problem, continued
Electric Field Strength
8. Evaluate your answer.
The electric field at point P is pointing away from the
charge q1, as expected, because q1 is a positive
charge and is larger than the negative charge q2.
The Electric Field
Electric Field Lines
• The number of electric
field lines is proportional
to the electric field
strength.
• Electric field lines are
tangent to the electric
field vector at any point.
The Electric Field
Conductors in Electrostatic Equilibrium
• The electric field is zero everywhere inside the
conductor.
• Any excess charge on an isolated conductor resides
entirely on the conductor’s outer surface.
• The electric field just outside a charged conductor is
perpendicular to the conductor’s surface.
• On an irregularly shaped conductor, charge tends to
accumulate where the radius of curvature of the
surface is smallest, that is, at sharp points.
Multiple Choice
1. In which way is the electric force similar to the
gravitational force?
A. Electric force is proportional to the mass of the
object.
B. Electric force is similar in strength to gravitational
force.
C. Electric force is both attractive and repulsive.
D. Electric force decreases in strength as the
distance between the charges increases.
Multiple Choice, continued
1. In which way is the electric force similar to the
gravitational force?
A. Electric force is proportional to the mass of the
object.
B. Electric force is similar in strength to gravitational
force.
C. Electric force is both attractive and repulsive.
D. Electric force decreases in strength as the
distance between the charges increases.
Multiple Choice, continued
2. What must the charges be for
A and B in the figure so that
they produce the electric
field lines shown?
F. A and B must both be
positive.
G. A and B must both be
negative.
H. A must be negative, and B
must be positive.
J. A must be positive, and B
must be negative.
Multiple Choice, continued
2. What must the charges be for
A and B in the figure so that
they produce the electric
field lines shown?
F. A and B must both be
positive.
G. A and B must both be
negative.
H. A must be negative, and B
must be positive.
J. A must be positive, and B
must be negative.
Multiple Choice, continued
3. Which activity does not produce the same results as
the other three?
A. sliding over a plastic-covered automobile seat
B. walking across a woolen carpet
C. scraping food from a metal bowl with a metal
spoon
D. brushing dry hair with a plastic comb
Multiple Choice, continued
3. Which activity does not produce the same results as
the other three?
A. sliding over a plastic-covered automobile seat
B. walking across a woolen carpet
C. scraping food from a metal bowl with a metal
spoon
D. brushing dry hair with a plastic comb
Multiple Choice, continued
4. By how much does the electric force between two
charges change when the distance between them is
doubled?
F. 4
G. 2
1
H.
2
1
J.
4
Multiple Choice, continued
4. By how much does the electric force between two
charges change when the distance between them is
doubled?
F. 4
G. 2
1
H.
2
1
J.
4
Multiple Choice, continued
Use the passage below to answer questions 5–6.
A negatively charged object is brought close to the
surface of a conductor, whose opposite side is then
grounded.
5. What is this process of charging called?
A. charging by contact
B. charging by induction
C. charging by conduction
D. charging by polarization
Multiple Choice, continued
Use the passage below to answer questions 5–6.
A negatively charged object is brought close to the
surface of a conductor, whose opposite side is then
grounded.
5. What is this process of charging called?
A. charging by contact
B. charging by induction
C. charging by conduction
D. charging by polarization
Multiple Choice, continued
Use the passage below to answer questions 5–6.
A negatively charged object is brought close to the
surface of a conductor, whose opposite side is then
grounded.
6. What kind of charge is left on the conductor’s
surface?
F. neutral
G. negative
H. positive
J. both positive and negative
Multiple Choice, continued
Use the passage below to answer questions 5–6.
A negatively charged object is brought close to the
surface of a conductor, whose opposite side is then
grounded.
6. What kind of charge is left on the conductor’s
surface?
F. neutral
G. negative
H. positive
J. both positive and negative
Multiple Choice, continued
Use the graph below to answer
questions 7–10. The graph shows
the electric field strength at different
distances from the center of the
charged conducting sphere of a Van
de Graaff generator.
7. What is the electric field
strength 2.0 m from the
center of the conducting
sphere?
A. 0 N/C
B. 5.0  102 N/C
C. 5.0  103 N/C
D. 7.2  103 N/C
Multiple Choice, continued
Use the graph below to answer
questions 7–10. The graph shows
the electric field strength at different
distances from the center of the
charged conducting sphere of a Van
de Graaff generator.
7. What is the electric field
strength 2.0 m from the
center of the conducting
sphere?
A. 0 N/C
B. 5.0  102 N/C
C. 5.0  103 N/C
D. 7.2  103 N/C
Multiple Choice, continued
Use the graph below to answer
questions 7–10. The graph shows
the electric field strength at different
distances from the center of the
charged conducting sphere of a Van
de Graaff generator.
8. What is the strength of
the electric field at the
surface of the
conducting sphere?
F. 0 N/C
G. 1.5  102 N/C
H. 2.0  102 N/C
J. 7.2  103 N/C
Multiple Choice, continued
Use the graph below to answer
questions 7–10. The graph shows
the electric field strength at different
distances from the center of the
charged conducting sphere of a Van
de Graaff generator.
8. What is the strength of
the electric field at the
surface of the
conducting sphere?
F. 0 N/C
G. 1.5  102 N/C
H. 2.0  102 N/C
J. 7.2  103 N/C
Multiple Choice, continued
Use the graph below to answer
questions 7–10. The graph shows
the electric field strength at different
distances from the center of the
charged conducting sphere of a Van
de Graaff generator.
9. What is the strength of
the electric field inside
the conducting sphere?
A. 0 N/C
B. 1.5  102 N/C
C. 2.0  102 N/C
D. 7.2  103 N/C
Multiple Choice, continued
Use the graph below to answer
questions 7–10. The graph shows
the electric field strength at different
distances from the center of the
charged conducting sphere of a Van
de Graaff generator.
9. What is the strength of
the electric field inside
the conducting sphere?
A. 0 N/C
B. 1.5  102 N/C
C. 2.0  102 N/C
D. 7.2  103 N/C
Multiple Choice, continued
Use the graph below to answer
questions 7–10. The graph shows
the electric field strength at different
distances from the center of the
charged conducting sphere of a Van
de Graaff generator.
10. What is the radius of
the conducting sphere?
F. 0.5 m
G. 1.0 m
H. 1.5 m
J. 2.0 m
Multiple Choice, continued
Use the graph below to answer
questions 7–10. The graph shows
the electric field strength at different
distances from the center of the
charged conducting sphere of a Van
de Graaff generator.
10. What is the radius of
the conducting sphere?
F. 0.5 m
G. 1.0 m
H. 1.5 m
J. 2.0 m
Short Response
11. Three identical
charges (q = +5.0 mC)
are along a circle with a
radius of 2.0 m at angles
of 30°, 150°, and 270°,
as shown in the
figure.What is the
resultant electric field at
the center?
Short Response, continued
11. Three identical
charges (q = +5.0 mC)
are along a circle with a
radius of 2.0 m at angles
of 30°, 150°, and 270°,
as shown in the
figure.What is the
resultant electric field at
the center?
Answer: 0.0 N/C
Short Response, continued
12. If a suspended object is attracted to another object
that is charged, can you conclude that the suspended
object is charged? Briefly explain your answer.
Short Response, continued
12. If a suspended object is attracted to another object
that is charged, can you conclude that the suspended
object is charged? Briefly explain your answer.
Answer: not necessarily; The suspended object might
have a charge induced on it, but its overall charge
could be neutral.
Short Response, continued
13. One gram of hydrogen contains 6.02  1023 atoms,
each with one electron and one proton. Suppose that
1.00 g of hydrogen is separated into protons and
electrons, that the protons are placed at Earth’s north
pole, and that the electrons are placed at Earth’s
south pole. Assuming the radius of Earth to be 6.38 
106 m, what is the magnitude of the resulting
compressional force on Earth?
Short Response, continued
13. One gram of hydrogen contains 6.02  1023 atoms,
each with one electron and one proton. Suppose that
1.00 g of hydrogen is separated into protons and
electrons, that the protons are placed at Earth’s north
pole, and that the electrons are placed at Earth’s
south pole. Assuming the radius of Earth to be 6.38 
106 m, what is the magnitude of the resulting
compressional force on Earth?
Answer: 5.12  105 N
Short Response, continued
14. Air becomes a conductor when the electric field
strength exceeds 3.0  106 N/C. Determine the
maximum amount of charge that can be carried by a
metal sphere 2.0 m in radius.
Short Response, continued
14. Air becomes a conductor when the electric field
strength exceeds 3.0  106 N/C. Determine the
maximum amount of charge that can be carried by a
metal sphere 2.0 m in radius.
Answer: 1.3  10–3 C
Extended Response
Use the information
below to answer
questions 15–18.
A proton, which has a
mass of 1.673  10–27 kg,
accelerates from rest in a
uniform electric field of
640 N/C. At some time
later, its speed is 1.2  106
m/s.
15. What is the magnitude
of the acceleration of
the proton?
Extended Response, continued
Use the information
below to answer
questions 15–18.
A proton, which has a
mass of 1.673  10–27 kg,
accelerates from rest in a
uniform electric field of
640 N/C. At some time
later, its speed is 1.2  106
m/s.
15. What is the magnitude
of the acceleration of
the proton?
Answer: 6.1  1010 m/s2
Extended Response, continued
Use the information
below to answer
questions 15–18.
A proton, which has a
mass of 1.673  10–27 kg,
accelerates from rest in a
uniform electric field of
640 N/C. At some time
later, its speed is 1.2  106
m/s.
16. How long does it take
the proton to reach this
speed?
Extended Response, continued
Use the information
below to answer
questions 15–18.
A proton, which has a
mass of 1.673  10–27 kg,
accelerates from rest in a
uniform electric field of
640 N/C. At some time
later, its speed is 1.2  106
m/s.
16. How long does it take
the proton to reach this
speed?
Answer: 2.0  10–5 s
Extended Response, continued
Use the information
below to answer
questions 15–18.
A proton, which has a
mass of 1.673  10–27 kg,
accelerates from rest in a
uniform electric field of
640 N/C. At some time
later, its speed is 1.2  106
m/s.
17. How far has it moved
in this time interval?
Extended Response, continued
Use the information
below to answer
questions 15–18.
A proton, which has a
mass of 1.673  10–27 kg,
accelerates from rest in a
uniform electric field of
640 N/C. At some time
later, its speed is 1.2  106
m/s.
17. How far has it moved
in this time interval?
Answer: 12 m
Extended Response, continued
Use the information
below to answer
questions 15–18.
A proton, which has a
mass of 1.673  10–27 kg,
accelerates from rest in a
uniform electric field of
640 N/C. At some time
later, its speed is 1.2  106
m/s.
18. What is its kinetic
energy at the later
time?
Extended Response, continued
Use the information
below to answer
questions 15–18.
A proton, which has a
mass of 1.673  10–27 kg,
accelerates from rest in a
uniform electric field of
640 N/C. At some time
later, its speed is 1.2  106
m/s.
18. What is its kinetic
energy at the later
time?
Answer: 1.2  10–15 J
Extended Response, continued
19. A student standing on a piece of insulating material
places her hand on a Van de Graaff generator. She
then turns on the generator. Shortly thereafter, her
hairs stand on end. Explain how charge is or is not
transferred in this situation, why the student is not
shocked, and what causes her hairs to stand up after
the generator is started.
Extended Response, continued
19. (See previous slide for question.)
Answer: The charge on the sphere of the Van de Graaff
generator is transferred to the student by means of
conduction. This charge remains on the student
because she is insulated from the ground. As there is
no path between the student and the generator and
the student and the ground by which charge can
escape, the student is not shocked. The
accumulation of charges of the same sign on the
strands of the student’s hair causes the strands to
repel each other and so stand on end.
Electric Charge
Charging By Induction
Electric Charge
Transfer of Electric Charge
The Electric Field
Electric Field Lines
Electric Potential
Electrical Potential Energy
• Electrical potential energy is potential energy
associated with a charge due to its position in an
electric field.
• Electrical potential energy is a component of
mechanical energy.
ME = KE + PEgrav + PEelastic + PEelectric
Electric Potential
Electrical Potential Energy, continued
• Electrical potential energy can be associated with a
charge in a uniform field.
• Electrical Potential Energy in a Uniform Electric Field
PEelectric = –qEd
electrical potential energy = –(charge)  (electric field strength) 
(displacement from the reference point in the direction of the field)
Electric Potential
Potential Difference
• Electric Potential equals the work that must be
performed against electric forces to move a charge
from a reference point to the point in question,
divided by the charge.
• The electric potential associated with a charge is the
electric energy divided by the charge:
PEelectric
V
q
Electric Potential
Potential Difference, continued
• Potential Difference equals the work that must be
performed against electric forces to move a charge
between the two points in question, divided by the
charge.
• Potential difference is a change in electric potential.
PEelectric
V 
q
change in electric potential energy
potential difference 
electric charge
Electric Potential
Potential Difference, continued
• The potential difference in a uniform field varies with
the displacement from a reference point.
• Potential Difference in a Uniform Electric Field
∆V = –Ed
potential difference = –(magnitude of the electric
field  displacement)
Electric Potential
Sample Problem
Potential Energy and Potential Difference
A charge moves a distance of 2.0 cm in the
direction of a uniform electric field whose
magnitude is 215 N/C.As the charge moves, its
electrical potential energy decreases by 6.9  10-19
J. Find the charge on the moving particle. What is
the potential difference between the two
locations?
Electric Potential
Sample Problem, continued
Potential Energy and Potential Difference
Given:
∆PEelectric = –6.9  10–19 J
d = 0.020 m
E = 215 N/C
Unknown:
q=?
∆V = ?
Electric Potential
Sample Problem, continued
Potential Energy and Potential Difference
Use the equation for the change in electrical potential
energy.
PEelectric = –qEd
Rearrange to solve for q, and insert values.
PEelectric
(–6.9  10 –19 J)
q–
–
Ed
(215 N/C)(0.020 m)
q  1.6  10 –19 C
Electric Potential
Sample Problem, continued
Potential Energy and Potential Difference
The potential difference is the magnitude of E times
the displacement.
V  –Ed  –(215 N/C)(0.020 m)
V  –4.3 V
Electric Potential
Potential Difference, continued
• At right, the electric potential at point A depends on
the charge at point B and
the distance r.
• An electric potential exists
at some point in an electric
field regardless of whether
there is a charge at that
point.
Electric Potential
Potential Difference, continued
• The reference point for potential difference near a
point charge is often at infinity.
• Potential Difference Between a Point at Infinity and a
Point Near a Point Charge
q
V  kC
r
potential difference = Coulomb constant 
value of the point charge
distance to the point charge
• The superposition principle can be used to calculate
the electric potential for a group of charges.
Capacitance
Capacitors and Charge Storage
• A capacitor is a device that is used to store electrical
potential energy.
• Capacitance is the ability of a conductor to store
energy in the form of electrically separated charges.
• The SI units for capacitance is the farad, F, which
equals a coulomb per volt (C/V)
Capacitance
Capacitors and Charge Storage, continued
• Capacitance is the ratio of charge to potential
difference.
Q
C
V
magnitude of charge on each plate
capacitance =
potential difference
Capacitance
Capacitors and Charge Storage, continued
• Capacitance depends on the size and shape of a
capacitor.
• Capacitance for a Parallel-Plate Capacitor in a
Vacuum
A
C  0
d
capacitance = permittivity of a vacuum 
area of one of the plates
distance between the plates
 0  permittivity of the medium  8.85  10 C /N  m
–12
2
Capacitance
Capacitors and Charge Storage, continued
• The material between a
capacitor’s plates can
change its capacitance.
• The effect of a dielectric
is to reduce the strength
of the electric field in a
capacitor.
Capacitance
Capacitors in Keyboards
Capacitance
Energy and Capacitors
• The potential energy stored in a charged capacitor
depends on the charge and the potential difference
between the capacitor’s two plates.
1
PEelectric  QV
2
electrical potential energy =
1
2
(charge on one plate)(final potential difference)
Capacitance
Sample Problem
Capacitance
A capacitor, connected to a 12 V battery, holds 36
µC of charge on each plate. What is the
capacitance of the capacitor? How much electrical
potential energy is stored in the capacitor?
Given:
Q = 36 µC = 3.6  10–5 C
∆V = 12 V
Unknown:
C=?
PEelectric = ?
Capacitance
Sample Problem, continued
Capacitance
To determine the capacitance, use the definition of
capacitance.
Q
3.6  10 –5 C
C

V
12 V
C  3.0  10 –6 F  3.0 µF
Capacitance
Sample Problem, continued
Capacitance
To determine the potential energy, use the
alternative form of the equation for the potential
energy of a charged capacitor:
PEelectric
PEelectric
1
 C( V )2
2
1
 (3.0  10 –6 F)(12 V)2
2
PEelectric  2.2  10 –4 J
Current and Resistance
Current and Charge Movement
• Electric current is the rate at which electric charges
pass through a given area.
I
electric current =
Q
t
charge passing through a given area
time interval
Current and Resistance
Drift Velocity
• Drift velocity is the the
net velocity of a charge
carrier moving in an
electric field.
• Drift speeds are
relatively small because
of the many collisions
that occur when an
electron moves through
a conductor.
Current and Resistance
Resistance to Current
• Resistance is the opposition presented to electric
current by a material or device.
• The SI units for resistance is the ohm (Ω) and is
equal to one volt per ampere.
• Resistance
V
I
potential difference
resistance 
current
R
Current and Resistance
Resistance to Current, continued
• For many materials resistance is constant over a
range of potential differences. These materials obey
Ohm’s Law and are called ohmic materials.
• Ohm’s low does not hold for all materials. Such
materials are called non-ohmic.
• Resistance depends on length, cross-sectional area,
temperature, and material.
Current and Resistance
Resistance to Current, continued
• Resistors can be used to control the amount of
current in a conductor.
• Salt water and perspiration lower the body's
resistance.
• Potentiometers have variable resistance.
Electric Power
Sources and Types of Current
• Batteries and generators supply energy to charge
carriers.
• Current can be direct or alternating.
– In direct current, charges move in a single
direction.
– In alternating current, the direction of charge
movement continually alternates.
Electric Power
Energy Transfer
• Electric power is the rate of conversion of electrical
energy.
• Electric power
P = I∆V
Electric power = current  potential difference
Electric Power
Energy Transfer, continued
• Power dissipated by a resistor
2
(

V
)
P  I V  I 2R 
R
• Electric companies measure energy consumed in
kilowatt-hours.
• Electrical energy is transferred at high potential
differences to minimize energy loss.
Multiple Choice
1. What changes would take place if the
electron moved from point A to point B in
the uniform electric field?
A. The electron’s electrical potential
energy would increase; its electric
potential would increase.
B. The electron’s electrical potential
energy would increase; its electric
potential would decrease.
C. The electron’s electrical potential
energy would decrease; its electric
potential would decrease.
D. Neither the electron’s electrical
potential energy nor its electric potential
would change.
Multiple Choice, continued
1. What changes would take place if the
electron moved from point A to point B in
the uniform electric field?
A. The electron’s electrical potential
energy would increase; its electric
potential would increase.
B. The electron’s electrical potential
energy would increase; its electric
potential would decrease.
C. The electron’s electrical potential
energy would decrease; its electric
potential would decrease.
D. Neither the electron’s electrical
potential energy nor its electric potential
would change.
Multiple Choice, continued
2. What changes would take place if the
electron moved from point A to point C in
the uniform electric field?
F. The electron’s electrical potential
energy would increase; its electric
potential would increase.
G. The electron’s electrical potential
energy would increase; its electric
potential would decrease.
H. The electron’s electrical potential
energy would decrease; its electric
potential would decrease.
J. Neither the electron’s electrical
potential energy nor its electric potential
would change.
Multiple Choice, continued
2. What changes would take place if the
electron moved from point A to point C in
the uniform electric field?
F. The electron’s electrical potential
energy would increase; its electric
potential would increase.
G. The electron’s electrical potential
energy would increase; its electric
potential would decrease.
H. The electron’s electrical potential
energy would decrease; its electric
potential would decrease.
J. Neither the electron’s electrical
potential energy nor its electric potential
would change.
Multiple Choice, continued
Use the following passage to answer questions 3–4.
A proton (q = 1.6  10–19 C) moves 2.0  10–6 m in the
direction of an electric field that has a magnitude of
2.0 N/C.
3. What is the change in the electrical potential energy
associated with the proton?
A. –6.4  10–25 J
B. –4.0  10–6 V
C. +6.4  10–25 J
D. +4.0  10–6 V
Multiple Choice, continued
Use the following passage to answer questions 3–4.
A proton (q = 1.6  10–19 C) moves 2.0  10–6 m in the
direction of an electric field that has a magnitude of
2.0 N/C.
3. What is the change in the electrical potential energy
associated with the proton?
A. –6.4  10–25 J
B. –4.0  10–6 V
C. +6.4  10–25 J
D. +4.0  10–6 V
Multiple Choice, continued
Use the following passage to answer questions 3–4.
A proton (q = 1.6  10–19 C) moves 2.0  10–6 m in the
direction of an electric field that has a magnitude of
2.0 N/C.
4. What is the potential difference between the proton’s
starting point and ending point?
F. –6.4  10–25 J
G. –4.0  10–6 V
H. +6.4  10–25 J
J. +4.0  10–6 V
Multiple Choice, continued
Use the following passage to answer questions 3–4.
A proton (q = 1.6  10–19 C) moves 2.0  10–6 m in the
direction of an electric field that has a magnitude of
2.0 N/C.
4. What is the potential difference between the proton’s
starting point and ending point?
F. –6.4  10–25 J
G. –4.0  10–6 V
H. +6.4  10–25 J
J. +4.0  10–6 V
Multiple Choice, continued
5. If the negative terminal of a 12 V battery is grounded,
what is the potential of the positive terminal?
A. –12 V
B. +0 V
C. +6 V
D. +12 V
Multiple Choice, continued
5. If the negative terminal of a 12 V battery is grounded,
what is the potential of the positive terminal?
A. –12 V
B. +0 V
C. +6 V
D. +12 V
Multiple Choice, continued
6. If the area of the plates of a parallel-plate capacitor is
doubled while the spacing between the plates is
halved, how is the capacitance affected?
F. C is doubled
G. C is increased by four times
H. C is decreased by 1/4
J. C does not change
Multiple Choice, continued
6. If the area of the plates of a parallel-plate capacitor is
doubled while the spacing between the plates is
halved, how is the capacitance affected?
F. C is doubled
G. C is increased by four times
H. C is decreased by 1/4
J. C does not change
Multiple Choice, continued
Use the following passage to answer questions 7–8.
A potential difference of 10.0 V exists across the
plates of a capacitor when the charge on each plate
is 40.0 µC.
7. What is the capacitance of the capacitor?
A. 2.00  10–4 F
B. 4.00  10–4 F
C. 2.00  10–6 F
D. 4.00  10–6 F
Multiple Choice, continued
Use the following passage to answer questions 7–8.
A potential difference of 10.0 V exists across the
plates of a capacitor when the charge on each plate
is 40.0 µC.
7. What is the capacitance of the capacitor?
A. 2.00  10–4 F
B. 4.00  10–4 F
C. 2.00  10–6 F
D. 4.00  10–6 F
Multiple Choice, continued
Use the following passage to answer questions 7–8.
A potential difference of 10.0 V exists across the
plates of a capacitor when the charge on each plate
is 40.0 µC.
8. How much electrical potential energy is stored in the
capacitor?
F. 2.00  10–4 J
G. 4.00  10–4 J
H. 2.00  10–6 J
J. 4.00  10–6 J
Multiple Choice, continued
Use the following passage to answer questions 7–8.
A potential difference of 10.0 V exists across the
plates of a capacitor when the charge on each plate
is 40.0 µC.
8. How much electrical potential energy is stored in the
capacitor?
F. 2.00  10–4 J
G. 4.00  10–4 J
H. 2.00  10–6 J
J. 4.00  10–6 J
Multiple Choice, continued
9. How long does it take 5.0 C of charge to pass
through a given cross section of a copper wire if I =
5.0 A?
A. 0.20 s
B. 1.0 s
C. 5.0 s
D. 25 s
Multiple Choice, continued
9. How long does it take 5.0 C of charge to pass
through a given cross section of a copper wire if I =
5.0 A?
A. 0.20 s
B. 1.0 s
C. 5.0 s
D. 25 s
Multiple Choice, continued
10. A potential difference of 12 V produces a current of
0.40 A in a piece of copper wire. What is the
resistance of the wire?
F. 4.8 Ω
G. 12 Ω
H. 30 Ω
J. 36 Ω
Multiple Choice, continued
10. A potential difference of 12 V produces a current of
0.40 A in a piece of copper wire. What is the
resistance of the wire?
F. 4.8 Ω
G. 12 Ω
H. 30 Ω
J. 36 Ω
Multiple Choice, continued
11. How many joules of energy are dissipated by a 50.0
W light bulb in 2.00 s?
A. 25.0 J
B. 50.0 J
C. 100 J
D. 200 J
Multiple Choice, continued
11. How many joules of energy are dissipated by a 50.0
W light bulb in 2.00 s?
A. 25.0 J
B. 50.0 J
C. 100 J
D. 200 J
Multiple Choice, continued
12. How much power is needed to operate a radio that
draws 7.0 A of current when a potential difference of
115 V is applied across it?
F. 6.1  10–2 W
G. 2.3  100 W
H. 1.6  101 W
J. 8.0  102 W
Multiple Choice, continued
12. How much power is needed to operate a radio that
draws 7.0 A of current when a potential difference of
115 V is applied across it?
F. 6.1  10–2 W
G. 2.3  100 W
H. 1.6  101 W
J. 8.0  102 W
Short Response
13. Electrons are moving from left to right in a wire. No
other charged particles are moving in the wire. In
what direction is the conventional current?
Short Response, continued
13. Electrons are moving from left to right in a wire. No
other charged particles are moving in the wire. In
what direction is the conventional current?
Answer: right to left
Short Response, continued
14. What is drift velocity, and how does it compare with
the speed at which an electric field travels through a
wire?
Short Response, continued
14. What is drift velocity, and how does it compare with
the speed at which an electric field travels through a
wire?
Answer: Drift velocity is the net velocity of a charge
carrier moving in an electric field. Drift velocities in a
wire are typically much smaller than the speeds at
which changes in the electric field propagate through
the wire.
Short Response, continued
15. List four factors that can affect the resistance of a
wire.
Short Response, continued
15. List four factors that can affect the resistance of a
wire.
Answer: length, cross-sectional area (thickness),
temperature, and material
Extended Response
16. A parallel-plate capacitor is made of two circular
plates, each of which has a diameter of 2.50  10–3
m. The plates of the capacitor are separated by a
space of 1.40  10–4 m.
a. Assuming that the capacitor is operating in a vacuum
and that the permittivity of a vacuum (0 = 8.85  10–
12 C2/N•m2) can be used, determine the capacitance
of the capacitor.
Extended Response, continued
16. A parallel-plate capacitor is made of two circular
plates, each of which has a diameter of 2.50  10–3
m. The plates of the capacitor are separated by a
space of 1.40  10–4 m.
a. Assuming that the capacitor is operating in a vacuum
and that the permittivity of a vacuum (0 = 8.85  10–
12 C2/N•m2) can be used, determine the capacitance
of the capacitor.
Answer: 3.10  10–13 F
Extended Response, continued
16. A parallel-plate capacitor is made of two circular
plates, each of which has a diameter of 2.50  10–3
m. The plates of the capacitor are separated by a
space of 1.40  10–4 m.
b. How much charge will be stored on each plate of the
capacitor when the capacitor’s plates are connected
across a potential difference of 0.12 V?
Extended Response, continued
16. A parallel-plate capacitor is made of two circular
plates, each of which has a diameter of 2.50  10–3
m. The plates of the capacitor are separated by a
space of 1.40  10–4 m.
b. How much charge will be stored on each plate of the
capacitor when the capacitor’s plates are connected
across a potential difference of 0.12 V?
Answer: 3.7  10–14 C
Extended Response, continued
16. A parallel-plate capacitor is made of two circular
plates, each of which has a diameter of 2.50  10–3
m. The plates of the capacitor are separated by a
space of 1.40  10–4 m.
c. What is the electrical potential energy stored in the
capacitor when fully charged by the potential
difference of 0.12 V?
Extended Response, continued
16. A parallel-plate capacitor is made of two circular
plates, each of which has a diameter of 2.50  10–3
m. The plates of the capacitor are separated by a
space of 1.40  10–4 m.
c. What is the electrical potential energy stored in the
capacitor when fully charged by the potential
difference of 0.12 V?
Answer: 2.2  10–15 J
Extended Response, continued
16. A parallel-plate capacitor is made of two circular
plates, each of which has a diameter of 2.50  10–3
m. The plates of the capacitor are separated by a
space of 1.40  10–4 m.
d. What is the potential difference between a point
midway between the plates and a point that is 1.10 
10–4 m from one of the plates?
Extended Response, continued
16. A parallel-plate capacitor is made of two circular
plates, each of which has a diameter of 2.50  10–3
m. The plates of the capacitor are separated by a
space of 1.40  10–4 m.
d. What is the potential difference between a point
midway between the plates and a point that is 1.10 
10–4 m from one of the plates?
Answer: 3.4  10–2 V
Extended Response, continued
16. A parallel-plate capacitor is made of two circular
plates, each of which has a diameter of 2.50  10–3
m. The plates of the capacitor are separated by a
space of 1.40  10–4 m.
e. If the potential difference of 0.12 V is removed from
the circuit and the circuit is allowed to discharge until
the charge on the plates has decreased to 70.7
percent of its fully charged value, what will the
potential difference across the capacitor be?
Extended Response, continued
16. A parallel-plate capacitor is made of two circular
plates, each of which has a diameter of 2.50  10–3
m. The plates of the capacitor are separated by a
space of 1.40  10–4 m.
e. If the potential difference of 0.12 V is removed from
the circuit and the circuit is allowed to discharge until
the charge on the plates has decreased to 70.7
percent of its fully charged value, what will the
potential difference across the capacitor be?
Answer: 8.5  10–2 V
Capacitance
Charging a Capacitor
Capacitance
A Capacitor With a Dielectric
Schematic Diagrams and
Circuits
Schematic Diagrams
• A schematic diagram
is a representation of a
circuit that uses lines to
represent wires and
different symbols to
represent components.
• Some symbols used in
schematic diagrams are
shown at right.
Schematic Diagrams and
Circuits
Electric Circuits
• An electric circuit is a set of electrical components
connected such that they provide one or more
complete paths for the movement of charges.
• A schematic diagram for a circuit is sometimes called
a circuit diagram.
• Any element or group of elements in a circuit that
dissipates energy is called a load.
Schematic Diagrams and
Circuits
Electric Circuits, continued
• A circuit which contains a complete path for electrons
to follow is called a closed circuit.
• Without a complete path, there is no charge flow and
therefore no current. This situation is called an open
circuit.
• A short circuit is a closed circuit that does not contain
a load. Short circuits can be hazardous.
Schematic Diagrams and
Circuits
Electric Circuits, continued
• The source of potential difference and electrical
energy is the circuits emf.
• Any device that transforms nonelectrical energy into
electrical energy, such as a battery or a generator, is
a source of emf.
• If the internal resistance of a battery is neglected, the
emf equals the potential difference across the
source’s two terminals.
Schematic Diagrams and
Circuits
Electric Circuits, continued
• The terminal voltage is the potential difference across
a battery’s positive and negative terminals.
• For conventional current, the terminal voltage is less
than the emf.
• The potential difference across a load equals the
terminal voltage.
Schematic Diagrams and
Circuits
Light Bulb
Resistors in Series or in Parallel
Resistors in Series
• A series circuit describes two or more components of
a circuit that provide a single path for current.
• Resistors in series carry the same current.
• The equivalent resistance can be used to find the
current in a circuit.
• The equivalent resistance in a series circuit is the
sum of the circuit’s resistances.
Req = R1 + R2 + R3…
Resistors in Series or in Parallel
Resistors in Series
Resistors in Series or in Parallel
Resistors in Series, continued
• Two or more resistors in
the actual circuit have the
same effect on the current
as one equivalent resistor.
• The total current in a
series circuit equals the
potential difference divided
by the equivalent
resistance.
V
I
Req
Resistors in Series or in Parallel
Sample Problem
Resistors in Series
A 9.0 V battery is
connected to four light
bulbs, as shown at
right. Find the
equivalent resistance
for the circuit and the
current in the circuit.
Resistors in Series or in Parallel
Sample Problem, continued
Resistors in Series
1. Define
Given:
∆V = 9.0 V
R1 = 2.0 Ω
R2 = 4.0 Ω
R3 = 5.0 Ω
R4 = 7.0 Ω
Unknown:
Req = ?
I=?
Diagram:
Resistors in Series or in Parallel
Sample Problem, continued
Resistors in Series
2. Plan
Choose an equation or situation: Because the
resistors are connected end to end, they are in
series. Thus, the equivalent resistance can be
calculated with the equation for resistors in series.
Req = R1 + R2 + R3…
The following equation can be used to calculate the
current.
∆V = IReq
Resistors in Series or in Parallel
Sample Problem, continued
Resistors in Series
2. Plan, continued
Rearrange the equation to isolate the unknown:
No rearrangement is necessary to calculate Req, but
∆V = IReq must be rearranged to calculate the
current.
V
I
Req
Resistors in Series or in Parallel
Sample Problem, continued
Resistors in Series
3. Calculate
Substitute the values into the equation and solve:
Req = 2.0 Ω + 4.0 Ω + 5.0 Ω + 7.0 Ω
Req = 18.0 Ω
Substitute the equivalent resistance value into the
equation for current.
V
9.0 V
I

 0.50 A
Req 18.0 Ω
Resistors in Series or in Parallel
Sample Problem, continued
Resistors in Series
4. Evaluate
For resistors connected in series, the equivalent
resistance should be greater than the largest
resistance in the circuit.
18.0 Ω > 7.0 Ω
Resistors in Series or in Parallel
Resistors in Series, continued
• Series circuits require all elements to conduct
electricity
• As seen below, a burned out filament in a string of
bulbs has the same effect as an open switch. Because
the circuit is no longer complete, there is no current.
Resistors in Series or in Parallel
Resistors in Parallel
• A parallel arrangement describes two or more components of a circuit that provide separate conducting
paths for current because the components are
connected across common points or junctions
• Lights wired in parallel have more than one path for
current. Parallel circuits do not require all elements to
conduct.
Resistors in Series or in Parallel
Resistors in Parallel
Resistors in Series or in Parallel
Resistors in Parallel, continued
• Resistors in parallel have the same potential
differences across them.
• The sum of currents in parallel resistors equals the
total current.
• The equivalent resistance of resistors in parallel can
be calculated using a reciprocal relationship
1
1
1
1



...
Req R1 R2 R3
Resistors in Series or in Parallel
Sample Problem
Resistors in Parallel
A 9.0 V battery is
connected to four
resistors, as shown at
right. Find the
equivalent resistance
for the circuit and the
total current in the
circuit.
Resistors in Series or in Parallel
Sample Problem, continued
Resistors in Parallel
1. Define
Given:
∆V = 9.0 V
R1 = 2.0 Ω
R2 = 4.0 Ω
R3 = 5.0 Ω
R4 = 7.0 Ω
Unknown:
Req = ?
I=?
Diagram:
Resistors in Series or in Parallel
Sample Problem, continued
Resistors in Parallel
2. Plan
Choose an equation or situation: Because both
sides of each resistor are connected to common
points, they are in parallel. Thus, the equivalent
resistance can be calculated with the equation for
resistors in parallel.
1
1
1
1



...
Req R1 R2 R3
The following equation can be used to calculate
the current.
∆V = IReq
Resistors in Series or in Parallel
Sample Problem, continued
Resistors in Parallel
2. Plan, continued
Rearrange the equation to isolate the unknown:
No rearrangement is necessary to calculate Req;
rearrange ∆V = IReq to calculate the total current
delivered by the battery.
V
I
Req
Resistors in Series or in Parallel
Sample Problem, continued
Resistors in Parallel
3. Calculate
Substitute the values into the equation and solve:
1
1
1
1
1
=
+
+
+
Req
2.0 Ω
4.0 Ω
5.0 Ω
7.0 Ω
1
0.50
0.25
0.20
0.14 1.09
=
+
+
+

Req
Ω
Ω
Ω
Ω
Ω
Req
1Ω
=
= 0.917 Ω
1.09
Resistors in Series or in Parallel
Sample Problem, continued
Resistors in Parallel
3. Calculate, continued
Substitute the equivalent resistance value into the
equation for current.
V
9.0 V
I

Req 0.917 Ω
I  9.8 A
Resistors in Series or in Parallel
Sample Problem, continued
Resistors in Parallel
4. Evaluate
For resistors connected in parallel, the equivalent
resistance should be less than the smallest
resistance in the circuit.
0.917 Ω < 2.0 Ω
Resistors in Series or in Parallel
Resistors in Series or in Parallel
Complex Resistor Combinations
Resistors Combined Both in Parallel and in
Series
• Many complex circuits can be understood by isolating
segments that are in series or in parallel and
simplifying them to their equivalent resistances.
• Work backward to find the current in and potential
difference across a part of a circuit.
Complex Resistor Combinations
Sample Problem
Equivalent Resistance
Determine the equivalent resistance of the complex
circuit shown below.
Complex Resistor Combinations
Sample Problem, continued
Equivalent Resistance
Reasoning
The best approach is to divide the circuit into groups
of series and parallel resistors. This way, the methods
presented in Sample Problems A and B can be used
to calculate the equivalent resistance for each group.
Complex Resistor Combinations
Sample Problem, continued
Equivalent Resistance
1. Redraw the circuit as a group of resistors along
one side of the circuit.
Because bends in a wire do not affect the circuit, they
do not need to be represented in a schematic
diagram. Redraw the circuit without the corners,
keeping the arrangement of the circuit elements the
same.
TIP: For now,
disregard the
emf source,
and work only
with the
resistances.
Complex Resistor Combinations
Sample Problem, continued
Equivalent Resistance
2. Identify components
in series, and calculate their equivalent
resistance.
Resistors in group (a) and
(b) are in series.
For group (a):
Req = 3.0 Ω + 6.0 Ω = 9.0 Ω
For group (b):
Req = 6.0 Ω + 2.0 Ω = 8.0 Ω
Complex Resistor Combinations
Sample Problem, continued
Equivalent Resistance
3. Identify components in
parallel, and calculate
their equivalent resistance.
Resistors in group (c) are in
parallel.
1
1
1
0.12 0.25 0.37





Req 8.0Ω 4.0Ω
1Ω
1Ω
1Ω
Req  2.7 Ω
Complex Resistor Combinations
Sample Problem, continued
Equivalent Resistance
4. Repeat steps 2 and 3 until
the resistors in the circuit
are reduced to a single
equivalent resistance.The
remainder of the resistors,
group (d), are in series.
For group (d):
Req  9.0Ω  2.7Ω  1.0Ω
Req  12.7Ω
Complex Resistor Combinations
Sample Problem
Current in and Potential Difference Across a Resistor
Determine the current in and potential difference
across the 2.0 Ω resistor highlighted in the figure
below.
Complex Resistor Combinations
Sample Problem, continued
Current in and Potential Difference Across a Resistor
Reasoning
First determine the total circuit current by reducing the
resistors to a single equivalent resistance. Then rebuild
the circuit in steps, calculating the current and potential
difference for the equivalent resistance of each group
until the current in and potential difference across the
2.0 Ω resistor are known.
Complex Resistor Combinations
Sample Problem, continued
Current in and Potential Difference Across a Resistor
1. Determine the equivalent resistance of the circuit.
The equivalent resistance of the circuit is 12.7 Ω, as
calculated in the previous Sample Problem.
Complex Resistor Combinations
Sample Problem, continued
Current in and Potential Difference Across a Resistor
2. Calculate the total current in the circuit.
Substitute the potential difference and equivalent
resistance in ∆V = IR, and rearrange the equation to
find the current delivered by the battery.
V
9.0 V
I

 0.71 A
Req 12.7 Ω
Complex Resistor Combinations
Sample Problem, continued
3. Determine a path from the
equivalent resistance found in
step 1 to the 2.0 Ω resistor.
Review the path taken to find the
equivalent resistance in the figure at
right, and work backward through this
path. The equivalent resistance for
the entire circuit is the same as the
equivalent resistance for group (d).
The center resistor in group (d) in turn
is the equivalent resistance for group
(c). The top resistor in group (c) is the
equivalent resistance for group (b),
and the right resistor in group (b) is
the 2.0 Ω resistor.
Complex Resistor Combinations
Sample Problem, continued
Current in and Potential Difference Across a Resistor
4. Follow the path determined in step 3, and calculate
the current in and potential difference across each
equivalent resistance. Repeat this process until the
desired values are found.
Complex Resistor Combinations
Sample Problem, continued
4. A. Regroup, evaluate, and calculate.
Replace the circuit’s equivalent resistance with group
(d). The resistors in group (d) are in series; therefore,
the current in each resistor is the same as the current
in the equivalent resistance, which equals 0.71 A.
The potential difference across the 2.7 Ω resistor in
group (d) can be calculated using ∆V = IR.
Given: I = 0.71 A R = 2.7 Ω
Unknown: ∆V = ?
∆V = IR = (0.71 A)(2.7 Ω) = 1.9 V
Complex Resistor Combinations
Sample Problem, continued
4. B. Regroup, evaluate, and calculate.
Replace the center resistor with group (c).
The resistors in group (c) are in parallel; therefore,
the potential difference across each resistor is the
same as the potential difference across the 2.7 Ω
equivalent resistance, which equals 1.9 V. The
current in the 8.0 Ω resistor in group (c) can be
calculated using ∆V = IR.
Given: ∆V = 1.9 V R = 8.0 Ω
Unknown: I = ?
V 1.9 V
I

 0.24 A
R
8.0 Ω
Complex Resistor Combinations
Sample Problem, continued
4. C. Regroup, evaluate, and calculate.
Replace the 8.0 Ω resistor with group (b).
The resistors in group (b) are in series; therefore, the
current in each resistor is the same as the current in
the 8.0 Ω equivalent resistance, which equals 0.24 A.
The potential difference
I  0.24 A
across the 2.0 Ω resistor can
be calculated using ∆V = IR.
Given: I = 0.24 A R = 2.0 Ω
Unknown: ∆V = ?
V  IR  (0.24 A)(2.0 Ω)
V  0.48 V
Multiple Choice
1. Which of the following is the correct term for a circuit
that does not have a closed-loop path for electron
flow?
A. closed circuit
B. dead circuit
C. open circuit
D. short circuit
Multiple Choice, continued
1. Which of the following is the correct term for a circuit
that does not have a closed-loop path for electron
flow?
A. closed circuit
B. dead circuit
C. open circuit
D. short circuit
Multiple Choice, continued
2. Which of the following is the correct term for a circuit
in which the load has been unintentionally bypassed?
F. closed circuit
G. dead circuit
H. open circuit
J. short circuit
Multiple Choice, continued
2. Which of the following is the correct term for a circuit
in which the load has been unintentionally bypassed?
F. closed circuit
G. dead circuit
H. open circuit
J. short circuit
Multiple Choice, continued
Use the diagram below to
answer questions 3–5.
3. Which of the circuit
elements contribute to
the load of the circuit?
A. Only A
B. A and B, but not C
C. Only C
D. A, B, and C
Multiple Choice, continued
Use the diagram below to
answer questions 3–5.
3. Which of the circuit
elements contribute to
the load of the circuit?
A. Only A
B. A and B, but not C
C. Only C
D. A, B, and C
Multiple Choice, continued
Use the diagram below to
answer questions 3–5.
4. Which of the following is
the correct equation for
the equivalent resistance of the circuit?
F. Req  RA  RB
1
1
1
G.


Req RA RB
H. Req  I V
1
1
1
1
J.



Req RA RB RC
Multiple Choice, continued
Use the diagram below to
answer questions 3–5.
4. Which of the following is
the correct equation for
the equivalent resistance of the circuit?
F. Req  RA  RB
1
1
1
G.


Req RA RB
H. Req  I V
1
1
1
1
J.



Req RA RB RC
Multiple Choice, continued
Use the diagram below to
answer questions 3–5.
5. Which of the following is
the correct equation for
the current in the
resistor?
A. I  I A  IB  IC
B. IB 
V
Req
C. IB  Itotal  I A
D. IB 
V
RB
Multiple Choice, continued
Use the diagram below to
answer questions 3–5.
5. Which of the following is
the correct equation for
the current in the
resistor?
A. I  I A  IB  IC
B. IB 
V
Req
C. IB  Itotal  I A
D. IB 
V
RB
Multiple Choice, continued
Use the diagram below to
answer questions 6–7.
6. Which of the following is
the correct equation for
the equivalent resistance of the circuit?
F. Req  RA  RB  RC
G.
1
1
1
1



Req RA RB RC
H. Req  I V
J. Req
 1
1 
 RA  


R
R
C 
 B
–1
Multiple Choice, continued
Use the diagram below to
answer questions 6–7.
6. Which of the following is
the correct equation for
the equivalent resistance of the circuit?
F. Req  RA  RB  RC
G.
1
1
1
1



Req RA RB RC
H. Req  I V
J. Req
 1
1 
 RA  


R
R
C 
 B
–1
Multiple Choice, continued
Use the diagram below to
answer questions 6–7.
7. Which of the following is
the correct equation for
the current in resistor B?
A. I  I A  IB  IC
B. IB 
V
Req
C. IB  Itotal  I A
VB
D. IB 
RB
Multiple Choice, continued
Use the diagram below to
answer questions 6–7.
7. Which of the following is
the correct equation for
the current in resistor B?
A. I  I A  IB  IC
B. IB 
V
Req
C. IB  Itotal  I A
VB
D. IB 
RB
Multiple Choice, continued
8. Three 2.0 Ω resistors are connected in series to a 12
V battery. What is the potential difference across
each resistor?
F. 2.0 V
G. 4.0 V
H. 12 V
J. 36 V
Multiple Choice, continued
8. Three 2.0 Ω resistors are connected in series to a 12
V battery. What is the potential difference across
each resistor?
F. 2.0 V
G. 4.0 V
H. 12 V
J. 36 V
Multiple Choice, continued
Use the following passage
to answer questions 9–11.
Six light bulbs are
connected in parallel to a
9.0 V battery. Each bulb
has a resistance of 3.0 Ω.
9. What is the potential
difference across each
bulb?
A. 1.5 V
B. 3.0 V
C. 9.0 V
D. 27 V
Multiple Choice, continued
Use the following passage
to answer questions 9–11.
Six light bulbs are
connected in parallel to a
9.0 V battery. Each bulb
has a resistance of 3.0 Ω.
9. What is the potential
difference across each
bulb?
A. 1.5 V
B. 3.0 V
C. 9.0 V
D. 27 V
Multiple Choice, continued
Use the following passage
to answer questions 9–11.
Six light bulbs are
connected in parallel to a
9.0 V battery. Each bulb
has a resistance of 3.0 Ω.
10. What is the current in
each bulb?
F. 0.5 A
G. 3.0 A
H. 4.5 A
J. 18 A
Multiple Choice, continued
Use the following passage
to answer questions 9–11.
Six light bulbs are
connected in parallel to a
9.0 V battery. Each bulb
has a resistance of 3.0 Ω.
10. What is the current in
each bulb?
F. 0.5 A
G. 3.0 A
H. 4.5 A
J. 18 A
Multiple Choice, continued
Use the following passage
to answer questions 9–11.
Six light bulbs are
connected in parallel to a
9.0 V battery. Each bulb
has a resistance of 3.0 Ω.
11. What is the total current
in the circuit?
A. 0.5 A
B. 3.0 A
C. 4.5 A
D. 18 A
Multiple Choice, continued
Use the following passage
to answer questions 9–11.
Six light bulbs are
connected in parallel to a
9.0 V battery. Each bulb
has a resistance of 3.0 Ω.
11. What is the total current
in the circuit?
A. 0.5 A
B. 3.0 A
C. 4.5 A
D. 18 A
Short Response
12. Which is greater, a battery’s terminal voltage or the
same battery’s emf? Explain why these two quantities
are not equal.
Short Response, continued
12. Which is greater, a battery’s terminal voltage or the
same battery’s emf? Explain why these two quantities
are not equal.
Answer: A battery’s emf is slightly greater than its
terminal voltage. The difference is due to the battery’s
internal resistance.
Short Response, continued
13. Describe how a short circuit could lead to a fire.
Short Response, continued
13. Describe how a short circuit could lead to a fire.
Answer: In a short circuit, the equivalent resistance of
the circuit drops very low, causing the current to be
very high. The higher current can cause wires still in
the circuit to overheat, which may in turn cause a fire
in materials contacting the wires.
Short Response, continued
14. Explain the advantage of wiring the bulbs in a string
of decorative lights in parallel rather than in series.
Short Response, continued
14. Explain the advantage of wiring the bulbs in a string
of decorative lights in parallel rather than in series.
Answer: If one bulb is removed, the other bulbs will still
carry current.
Extended Response
15. Using standard symbols for circuit elements, draw a
diagram of a circuit that contains a battery, an open
switch, and a light bulb in parallel with a resistor. Add
an arrow to indicate the direction of current if the
switch were closed.
Extended Response, continued
15. Using standard symbols for circuit elements, draw a
diagram of a circuit that contains a battery, an open
switch, and a light bulb in parallel with a resistor. Add
an arrow to indicate the direction of current if the
switch were closed.
Answer:
Extended Response, continued
Use the diagram below to
answer questions 16–17.
16. For the circuit shown,
calculate the following:
a. the equivalent
resistance of the circuit
b. the current in the light
bulb.
Show all your work for
both calculations.
Extended Response, continued
Use the diagram below to
answer questions 16–17.
16. For the circuit shown,
calculate the following:
a. the equivalent
resistance of the circuit
b. the current in the light
bulb.
Show all your work for
both calculations.
Answer: a. 4.2 Ω b. 2.9 A
Extended Response, continued
Use the diagram below to
answer questions 16–17.
17. After a period of time,
the 6.0 Ω resistor fails
and breaks. Describe
what happens to the
brightness of the bulb.
Support your answer.
Extended Response, continued
Use the diagram below to
answer questions 16–17.
17. Answer: The bulb will
grow dim. The loss of
the 6.0 Ω resistor
causes the equivalent
resistance of the circuit
to increase to 4.5 Ω. As
a result, the current in
the bulb drops to 2.7 A,
and the brightness of
the bulb decreases.
Extended Response, continued
18. Find the current in and potential difference across
each of the resistors in the following circuits:
a. a 4.0 Ω and a 12.0 Ω resistor wired in series with a
4.0 V source.
b. a 4.0 Ω and a 12.0 Ω resistor wired in parallel with a
4.0 V source.
Show all your work for each calculation.
Extended Response, continued
18. Find the current in and potential difference across
each of the resistors in the following circuits:
a. a 4.0 Ω and a 12.0 Ω resistor wired in series with a
4.0 V source.
b. a 4.0 Ω and a 12.0 Ω resistor wired in parallel with a
4.0 V source.
Show all your work for each calculation.
Answers: a. 4.0 Ω: 0.25 A, 1.0 V
12.0 Ω: 0.25 A, 3.0 V
b. 4.0 Ω: 1.0 A, 4.0 V
12.0 Ω: 0.33 A, 4.0 V
Extended Response, continued
19. Find the current in and potential difference across
each of the resistors in the following circuits:
a. a 150 Ω and a 180 Ω resistor wired in series with a
12 V source.
b. a 150 Ω and a 180 Ω resistor wired in parallel with a
12 V source.
Show all your work for each calculation.
Extended Response, continued
19. Find the current in and potential difference across
each of the resistors in the following circuits:
a. a 150 Ω and a 180 Ω resistor wired in series with a
12 V source.
b. a 150 Ω and a 180 Ω resistor wired in parallel with a
12 V source.
Show all your work for each calculation.
Answer: a.150 Ω: 0.036 A, 5.4 V
180 Ω: 0.036 A, 6.5 V
b. 150 Ω: 0.080 A, 12 V
180 Ω: 0.067 A, 12 V
Schematic Diagrams and
Circuits
Diagram Symbols